Solution:

Note that $a = b = c = 12$ is a solution. Now, fix $c = 12$, the original equation becomes

$$\begin{array}{c} \{a^{2} + b^{2} + 144 + (a + b + 12)^{2} = 12ab\} \\ \{\Rightarrow 2a^{2} + (24 - 10b)a + (2b^{2} + 24b + 288) = 0\} \end{array} \quad (1)$$

We see that by Vieta's theorem, taking (1) as a polynomial in $a$, there are 2 solutions of $a$, adding up to $5b - 12$.

Thus, if $a \leq b$ then we have a new set of solutions: $(5b - 12 - a, b, 12)$, which by symmetry gives $(b, 5b - 12 - a, 12)$ as a bigger set of solutions in $a, b$ (non-strict in $a$, strict in $b$ as $5b - 12 - a \geq 5b - b - b = 3b > b$) when $a, b \geq 12$.

Since we can repeat this process infinitely, as $a, b$ increases they always fulfill the requirement that $a, b \geq 12$, thus we can generate infinitely many solutions.